Solution
A Hand with a Void
Show the problem again
A 13-card bridge hand is dealt uniformly at random from a standard 52-card deck. A void is a suit with no cards in the hand. What is the probability that the hand has at least one void? Give your answer as a decimal rounded to four places (the exact reduced fraction is also accepted).
Worked solution
The answer is about 0.0511 (exactly 1621364909/31750677980). Let V_s be the event that suit s is void. A hand avoiding a given suit is drawn from the other 39 cards, so P(V_s) = C(39,13)/C(52,13). Two given suits are both void with probability C(26,13)/C(52,13), and three are void only when the hand is the entire fourth suit: C(13,13)/C(52,13). Four voids are impossible. By inclusion–exclusion, P(some void) = [4·C(39,13) − 6·C(26,13) + 4·C(13,13)] / C(52,13) ≈ 0.0511. The first term alone gives about 0.0512. The correction is tiny because double voids need the hand to sit entirely within two suits. Roughly one bridge hand in twenty has a void.
Source: Classic bridge-probability exercise in wide circulation; no traceable original source. Statement written for AxiomIQ.